Problem: Determine the number of positive integers $a$ less than $12$ such that the congruence $ax\equiv 1\pmod{12}$ has a solution in $x$.
Solution: The given congruence has a solution if and only if $a$ is invertible modulo $12$ since the congruence implies that $a,x$ are inverses of each other modulo $12$. In other words, $\gcd(12,a)=1$. The only such positive $a$ less than $12$ are $1,5,7,11$. So the number of possible values of $a$ is $\boxed{4}$.